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A125269 Minimal number of states required for a 2-symbol, 5-tuple Turing machine that takes n steps on an initially blank tape before halting. 1
1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 4, 4, 5, 5, 4, 5, 5, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
If BB(n) = A060843(n), then a(BB(n)) = n and that is the last occurrence of n in this sequence. a(n) will not become monotonic; if it did, we could compute BB(n), since a(n) is computable. a(n) diverges properly, but does so very slowly. The terms with values 1,2,3 were computed by exhaustive search. The terms with value 4 were inferred from knowing that they are greater than 3 and from the observation that for all n, a(n+1) <= a(n) + 1 (an easy construction). Using exhaustive search, may be able to extend the sequence to (most of) the terms up to and a bit beyond a(107) = 4, but going much further would likely require a more sophisticated method (see A052200).
If BB(n) = A060843(n), then a(BB(n)) = n and that is the last occurrence of n in this sequence. a(n) will not become monotonic; if it did, we could compute BB(n), since a(n) is computable. a(n) diverges properly, but does so very slowly. a(n+1) <= a(n) + 1 (an easy construction). - Martin Fuller, Feb 14 2007
LINKS
CROSSREFS
Sequence in context: A105517 A230774 A056813 * A077430 A105513 A004233
KEYWORD
nice,nonn
AUTHOR
Dustin Wehr (robert.wehr(AT)mail.mcgill.ca), Jan 16 2007, Jan 28 2007
EXTENSIONS
More terms from Martin Fuller, Feb 14 2007
STATUS
approved

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Last modified April 23 15:20 EDT 2024. Contains 371916 sequences. (Running on oeis4.)